Identify and describe the difference between: an internet connected computer's numerical address and the human readable name for that computer.

Some of your friends have gotten into the burgeoning field of time-series data mining, in which one looks for patterns in sequences of events that occur over time. purchases at stock exchanges"what's being bought" are one source of data with a natural ordering in time. given a long sequence s of such events, your friends want an efficient way to detect certain "patterns" in them"for example, they may want to know if the four events buy yahoo, buy ebay, buy yahoo, buy oracle occur in this sequence s, in order but not necessarily consecutively. they begin with a collection of possible events (e.g., the possible transactions) and a sequence s of n of these events. a given event may occur multiple times in s (e.g., yahoo stock may be bought many times in a single sequence s). we will say that a sequence s is a subsequence of s if there is a way to delete certain of the events from s so that the remaining events, in order, are equal to the sequence s . so, for example, the sequence of four events above is a subsequence of the sequence buy amazon, buy yahoo, buy ebay, buy yahoo, buy yahoo, buy oracle their goal is to be able to dream up short sequences and quickly detect whether they are subsequences of s. so this is the problem they pose to you: give an algorithm that takes two sequences of events"s of length m and s of length n, each possibly containing an event more than once"and decides in time o(m + n) whether s is a subsequence of s.

Someone with this coz i don’t really know what i can choose, just pick whatever u want. homework - you need to choose a website that you like or use frequently. you must visit the website and discuss 6 different features/parts/aspects of the website that you think makes it good. (100 words)

What is the worst-case complexity of the maxrepeats function? assume that the longest string in the names array is at most 25 characters wide (i.e., string comparison can be treated as o( class namecounter { private: int* counts; int nc; string* names; int nn; public: namecounter (int ncounts, int nnames); int maxrepeats() const; }; int namecounter: : maxrepeats () { int maxcount = 0; for (int i = 0; i < nc; ++i) { int count = 1; for (int j = i+1; j < nc; ++j) { if (names[i] == names[j]) ++count; } maxcount = max(count, maxcount); } return maxcount; }